# How to do it theoretically ? ( A very interesting doubt !)

I just spotted a good question in one of my physics books, please try to answer it!

Show that it is not possible for a $photon$ to be completely absorbed by a free $electron$.

Note by Swapnil Das
4 years, 10 months ago

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By conservation of momentum

$\displaystyle \dfrac{h}{\lambda } = m_e v$

$\implies$$\displaystyle \lambda = \dfrac{h}{m_e v}$ ........ (1)

Where 'v' is velocity of electron after absorbing photon and $m_e$ mass of electron

By conservation of energy

$\displaystyle \dfrac{hc}{\lambda } = \frac{1}{2} m_e v^2$ ........ (2)

Substituting value of $\lambda$ from equation (1) in (2)

$\displaystyle m_e vc = \frac{1}{2} m_e v^2$

From here we get

$\displaystyle v = 0$

Or

$\displaystyle v = 2c$

For the first one absorbing a photon and remain stationary is not possible (here in the case of "free" electron)

And clearly second one is not at all possible.

- 4 years, 10 months ago

why $v=2c$ is impossible ?

Also what is your status mean ? " 272 worth anything ? " @Krishna Sharma

- 4 years, 10 months ago

We define mass of moving particle with velocity $v$ as

$\displaystyle m = \dfrac{m_0}{\sqrt{1 - \frac{v^2}{c^2} }}$

Where $m_0$ is the rest mass of the particle.

It is not possible for a particle(matter) to attain speed greater than or equal to speed of light because in that case mass of particle will not be defined(or imaginary).

- 4 years, 10 months ago

Ahh yes , Thanks ! Actually I did not know anything about relativistic mechanics ...

Okay if It is not define , what happen , what occure that we can't define practically particle having speed greater than light ?

I guess whole mass will convert into Energy .. Am I right ?

Still you did not tell that what your status mean ? Is it your JEE _ Mains marks ?

- 4 years, 10 months ago

Nothing can exceed the velocity of light in accordance to Einstein's Special Theory of Relativity, except $tachyons$, according modern theoretical particle physics.

- 4 years, 10 months ago

In $\left(2\right)$, shouldn't you have $hc/\lambda + m_e(0)c^2 = \frac12 m_e(v)v^2$?

Staff - 4 years, 10 months ago

Hi Swapnil,

I don't know how far along you are in physics, so I will be explicit about all my assumptions.

Let's consider a photon and a free electron traveling along in space.

The initial energy of this system is given by the sum of the energy of the photon and the electron:

$E_i = E_\gamma + E_{i}$

and the final energy is given by $E_f = E_{f}$ (since the photon no longer exists after absorption)

$E_\gamma = E_{f} - E_{i}$

Conservation of momentum should hold as well, and we have $E_\gamma/c + p_{i}= p_{f}$ , or

$E_\gamma = c\left(p_{f}-p_{i}\right)$

since the momentum of a photon is given by its energy divided by the speed of light. This was shown experimentally by measuring the deflection of very thing gold sheets by incoming light beams.

If we set the two forms for $E_\gamma$ equal, we find

$p_{f}-p_{i} = \frac{E_f - E_i}{c}$

Now, the kinetic energy of a moving particle is given by $E(p) = \sqrt{{m_{0}}^2c^4 + p^2c^2}$ which is a convex function of the momentum $p$, i.e. $\displaystyle\frac{E(p_f) - E(p_i)}{c} \geq p_f - p_i$ with equality holding when $p_f=p_i$.

Therefore it is impossible to have $c\left(p_{f}-p_{i}\right) = E_{f} - E_{i}$, except when $p_i = p_f$, in other words when there is no photon absorbed.

Staff - 4 years, 10 months ago

Thank You sir for giving your time to comment. Just a thing I wanted to ask, is $E = mc^{2}$ incomplete? I guess for that reason alone you used the extended form of Einstein's Equation with relativistic momentum.

- 4 years, 10 months ago

I originally started with $E=mc^2$ but found the argument less straightforward than the above. It isn't incomplete, it's just that people often use it one way or another without specifying what they're really talking about. In many places, you'll see $E=mc^2$ where the person really means $E=m_0c^2$, so they're talking about the rest energy, but they drop the zero subscript.

If we really mean $mc^2$, i.e. $m$ is not the rest mass, but is the relativistic mass, then $E=mc^2$ is the whole story. To see this, consider:

\begin{aligned} E &= mc^2 \\ &= \frac{m_0}{\sqrt{1-v^2/c^2}}c^2 \\ E^2 &= m_0^2\frac{c^4}{1-v^2/c^2} \\ E^2 &= m_0^2c^4 + E^2v^2/c^2 \end{aligned}

Now, using $E=mc^2$ on the right side again, we get

$E^2 = m_0^2c^4 + m^2v^2c^2$

But $p = mv$, so finally we have

$E^2 = m_0^2c^4+p^2c^2$

Staff - 4 years, 10 months ago

Sir, I am an aspiring Theoretical physicist. Could you please advice me some techniques which would make me even stronger in physics? Thank You!

- 4 years, 10 months ago

If you want to be a theoretical physicist I would start thinking about problems you'd really like to know the answers to. There are a huge number of things to potentially study in theoretical physics, from particles, to the cosmos, to the physics of living systems. But no matter problems what you end up interested in, I would suggest learning your mechanics from Kleppner and Kolenkov.

Staff - 4 years, 10 months ago

- 4 years, 10 months ago

Sir, I just saw the book suggested by you and found it really helpful. Currently I am a ninth grader. Shall I learn trigonometry and calculus first and then proceed to the book? Also, Please tell me how to learn calculus at this age of 14 , or if any book can help me. Thanks!

- 4 years, 10 months ago

Show that total relativistic energy and momentum are not simultaneously conserved if such a thing happens. Since these laws cannot be violated it is not possible for a electron to completely absorb a photon.

- 4 years, 10 months ago

@Sudeep Salgia @Nishant Rai Thank U for commenting!

- 4 years, 10 months ago

- 4 years, 10 months ago

I have no clue.

But it seems they have figured it out, in terms of energy and momentum conservations.

- 4 years, 10 months ago

- 4 years, 10 months ago

OK if there are no answers I like by tonight, I'll weigh in! Thanks for asking me @Swapnil Das

Staff - 4 years, 10 months ago